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Thread: Hydrogen-Oxygen Reaction In Different States

  1. #1 Hydrogen-Oxygen Reaction In Different States 
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    So, in modern rocketry, specifically in orbital lift systems, the highest efficiency existing propulsive system is a LOX-LH2 rocket.

    Using this, an attempt to make an SSTO, which would revolutionize the entire field, barely failed, because with current material limits and constraints, LOX-LH2 just isn't efficient enough. The reaction isn't energetic enough. The tantalizing thing is, it's just barely not energetic enough. The measure of a rocket's efficiency, which directly comes from how energetic the reaction is, is specific impulse (ISP). LOX-LH2 rockets get a ISP of ~450, give or take a few. I've seen analysis before where an ISP of 470 was needed to create a working SSTO with current materials.

    So, go easy, I just got this from a 12th grade honors chemistry class, but a solid state reactant would make the reaction more energetic, yes? And since a more energetic reaction means just about a proportionally higher ISP, that golden ISP of 470 might be attainable if either the LOX or LH2 was a solid, instead?

    Rockets exist with one reactant in a liquid state, and the other in a solid state, these are hybrid rockets.

    So, these are my questions:

    A) How much more energetic (i.e, how much more energy does the reaction produce) is a solid oxygen + liquid hydrogen or liquid oxygen + solid hydrogen reaction compared to a liquid oxygen + liquid hydrogen reaction?

    B) If either one is more than 5% more energetic than the liquid-liquid reaction, then from my basic understanding it should give an ISP of 470 or more. If this is the case, why haven't engines been built that use one reactant as a solid?


    A little bit of an apology... It's a cross-disciplinary question that belongs both in the "Chemistry" and "Mechanical, Structure, and Chemical Engineering" forums. Perhaps after question (A) is answered it could be moved to the engineering forum?


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  3. #2  
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    I think hydrogen gasses and liquid can't go solid because of its quantum property. Heinsenberg uncertainty stated that hydrogen atom must obey to somekind of "minimum velocity" which prevent hydrogen liquid from freezing. So it is nature's law that prevent us from using hydrogen solid rocket.


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    Odd...
    Slush hydrogen - Wikipedia, the free encyclopedia
    Solid hydrogen - Wikipedia, the free encyclopedia

    I thought finding the energy released from a reaction was a simple problem, but still no response to question (a)

    To re-state it...

    (liquid) H2 + (liquid) O2 -> 2(H2O) + ??? kJ

    (solid) H2 + (liquid) O2 -> 2(H2O) + ??? kJ

    (liquid) H2 + (solid) O2 -> 2(H2O) + ??? kJ

    Thanks in advance to anyone that will take up the challenge.
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  5. #4  
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    Whenever you perform a combustion reaction it is the heat of combustion that provides the energy for propulsion. For liquid fuels the net heat energy available for propulsion is going to be the heat of reaction minus the latent heat required to vaporize the liquid and heat it up to reaction temperature. If you use solid hydrogen you also have to subtract the latent heat of fusion to melt the hydrogen and the sensible heat to raise it from around 14 degrees K to its vaporization temperature. Therefore I would expect that solid hydrogen or solid oxygen would provide less thrust energy per kilogram than using the liquid form. In effect, some of the solid fuel will be combusted just to preheat itself. You could obtain the same thrust but would have to carry more fuel to get it.
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    Quote Originally Posted by Bunbury View Post
    Whenever you perform a combustion reaction it is the heat of combustion that provides the energy for propulsion. For liquid fuels the net heat energy available for propulsion is going to be the heat of reaction minus the latent heat required to vaporize the liquid and heat it up to reaction temperature. If you use solid hydrogen you also have to subtract the latent heat of fusion to melt the hydrogen and the sensible heat to raise it from around 14 degrees K to its vaporization temperature. Therefore I would expect that solid hydrogen or solid oxygen would provide less thrust energy per kilogram than using the liquid form. In effect, some of the solid fuel will be combusted just to preheat itself. You could obtain the same thrust but would have to carry more fuel to get it.
    Is the loss of thermal energy during vaporization because the volume changes drastically? If so, then no. Because at it's core thermal (in this case chemical) rocket engines work by introducing heat into a fluid, which causes it to expand, and the expansion creates velocity, not heat. If exhaust from a certain propellant expanded enough to cool itself to room temperature, then it would produce far more thrust than if it only expanded until it cooled to several thousand degrees. There's a small chance I may be wrong there, but I'm willing to bet that's correct, since really the only thing that matters is velocity of the exhaust, and velocity comes from the propellant's expansion, done in a nozzle, to direct all of the expansion in one direction.

    Interesting point, though, I haven't thought of heat of fusion or during it's liquid state. Heating it up to vaporization would lose a lot of energy that could've otherwise been spent expanding a gas.
    (LH2 kept in the Shuttle's External Tank is kept at about 33 psi, at -252.8 *C, compared to LH2's boiling point of -252.87 *C.)

    This may be counteracted to some small, incomplete extent, by it's change in density from a solid to a liquid, from 0.086 g/cm^3 as a solid, to 0.07 g/cm^3 as a liquid at melting point, to 0.07099 g/cm^3 as a liquid at boiling point. Also the temperature range from melting to boiling is 14 K - 20 K, and the heat of vaporization, 0.904 kJ/mole is much greater than it's heat of fusion, 0.117 kJ/mole.

    The ultimate fact won't come from expectations or guessing, though. What needs to be done is some number-crunching on how much extra energy will come from reacting them where one or the other is a solid, then subtracting the heat of fusion and heat to go from 14K-20K, then compare that to the energy from a pure-liquid reaction.
    Heat of vaporization should be accounted for in both. The energy used for propulsion will be the energy imparted on the gas. Though it's less than 1 kJ/mole, so it probably won't be significant.

    ---

    But once again, my formal knowledge of chemistry is very limited. Isn't calculating the energy of a reaction somewhat simple? I really don't know, but I guess it involves something to do with the specific heat capacity of the substance in each state, and the products and reactants? Can someone just crunch the numbers please?

    (liquid) H2 + (liquid) O2 -> 2(H2O) + ??? kJ

    (solid) H2 + (liquid) O2 -> 2(H2O) + ??? kJ

    (liquid) H2 + (solid) O2 -> 2(H2O) + ??? kJ

    Or if it's not too complicated, at least explain how to do it?
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    Is the loss of thermal energy during vaporization because the volume changes drastically?
    There is no loss of energy. Melting and vaporization are endothermic meaning energy is absorbed by the material from its surroundings. The surroundings in this case are hot gases which donate some of their heat to melting and vaporizing the fuel so the flame temperature is lower than it would be if you started with gaseous fuel. So for the same amount of thrust you will need to burn the least fuel if you start with gaseous fuel, more if you start with liquid and still more if you start with solid.

    how much extra energy will come from reacting them where one or the other is a solid
    Solid fuel rocket fuel is a mixture of fuel and oxidant. If you mix liquid oxygen and liquid hydrogen with the intent of freezing the mixture I would imagine you would have an explosion rather than a controlled burn, but I don't know that for a fact.
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    (liquid) H2 + (liquid) O2 -> 2(H2O) + ??? kJ

    (solid) H2 + (liquid) O2 -> 2(H2O) + ??? kJ

    (liquid) H2 + (solid) O2 -> 2(H2O) + ??? kJ
    Have you tried Googling for "Entalphy"? Isn't that for calculating heat?
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    Quote Originally Posted by Bunbury View Post
    There is no loss of energy. Melting and vaporization are endothermic meaning energy is absorbed by the material from its surroundings. The surroundings in this case are hot gases which donate some of their heat to melting and vaporizing the fuel so the flame temperature is lower than it would be if you started with gaseous fuel. So for the same amount of thrust you will need to burn the least fuel if you start with gaseous fuel, more if you start with liquid and still more if you start with solid.
    I meant loss of heat, not energy

    True, but isn't it also true that more energy comes out of the reaction if the reactants are in a more dense state? i.e., although heat is lost due to heat of vaporization, two liquid compounds mixing produces more energy than the same compounds as a gas?

    Quote Originally Posted by Bunbury View Post
    Solid fuel rocket fuel is a mixture of fuel and oxidant. If you mix liquid oxygen and liquid hydrogen with the intent of freezing the mixture I would imagine you would have an explosion rather than a controlled burn, but I don't know that for a fact.
    There's really no difference in-between a controlled burn and an explosion. A rocket is a controlled explosion. The issue you're talking about is that the Burn Rate [see link] would be tremendously high. This would mean burning all of the propellant in moments, which would result in an unfeasibly high chamber pressure and acceleration, unsuitable for building a rocket.

    The idea wasn't for a solid rocket like you've described, but to have only one of the propellants as a solid, and the other as a liquid, this is a Hybrid Rocket setup [see link]. I explained that in the OP . One reactant is kept as a solid, the other as a liquid. The solid is kept in the combustion chamber and reacts with the liquid, which is injected into the chamber. It is throttle-able and because you can change the oxidizer or fuel flow/mixture rate, you can control the burn rate.
    (IIRC, the oxidizer is usually the liquid due to some chemical reaction issues that result in less efficiency when it's the solid)
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    Quote Originally Posted by msafwan View Post
    Have you tried Googling for "Entalphy"? Isn't that for calculating heat?
    "Enthalpy", I've done a little bit of armchair research, from what it looks like I'm wrong about solids reacting producing more energy than liquids or gasses reacting. So it looks like the entire basis for the thread/idea is wrong

    I got the idea from some notes that said the states of reactants effect the change of enthalpy of the reaction, I asked the teacher and he said that was correct, solids giving more heat in a reaction. Though I'm beginning to suspect either he doesn't actually know or there was a miscommunication. High school teachers...
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  11. #10  
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    Quote Originally Posted by Eagle1_Division View Post
    There's really no difference in-between a controlled burn and an explosion. A rocket is a controlled explosion.
    A controlled explosion is an oxymoron (except perhaps in the sense of bomb disposal where the meaning of controlled is somewhat different).

    The idea wasn't for a solid rocket like you've described, but to have only one of the propellants as a solid, and the other as a liquid
    OK, then you have to include the latent heat and sensible heat to raise the solid and the liquid to the gaseous state in your calculations.
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